How to evaluate this inequality in a priori estimate of strong solution of SDEs

Question

I'm studying the existence and uniqueness of stochastic differential equations from https://link.springer.com/book/10.1007/978-1-4939-7256-2 book by Jianfeng Zhang. I'm trying to go through and understand how he proved the following theorem

Theorem: Assume

1. $b$ and $\sigma$ are $\mathbb{F}$-measurable.
2. $b$ and $\sigma$ are uniformly Lipschitz continuous in $x$ with a Lipschitz constant $L$, i.e.,
   $$|b(t,x_1) - b(t, x_2)| + |\sigma(t,x_1) - \sigma(t, x_2)\leq L|x_1 - x_2|, \quad \mbox{for all } \ x_1,x_2 \in \mathbb{R}.$$
3. $x_0 \in \mathbb{L}^2(\mathcal{F})$ := {$\mathbb{R}$-valued $\mathcal{F}$- measurable random variables $X$ s.t. $\| X \|_2 := \mathbb{E}[|X|^2]<\infty$}

$b(t,0) \in \mathbb{L}^{1,2}(\mathbb{F})$ := {$\mathbb{R}$-valued $\mathbb{F}$- measurable processes $X$ s.t. $\| X \|_{1,2} := \mathbb{E}[(\int_0^T|X|)^2]<\infty$ }

$\sigma(t,0) \in \mathbb{L}^{2}(\mathbb{F})$ := {$\mathbb{R}$-valued $\mathbb{F}$- measurable processes $X$ s.t. $\| X \|_{2} := \mathbb{E}[\int_0^T|X|^{2}]<\infty$ }

4. linear growth condition $$|b(t, x)| \leq |b(t,0)| + L |x|, \quad |\sigma(t, x)| \leq |\sigma(t,0)| + L |x|$$

5. $X \in \mathbb{L}^{2}(\mathbb{F}) $ is a solution to the SDE $$X_t = x_0 + \int_0^t b(s,X_s)ds + \int_0^t \sigma(s,X_s)dW_s$$.

Then $X \in \mathbb{S}^2(\mathbb{F})$ :={$\mathbb{R}$-valued $\mathbb{F}$- measurable random processes $X$ s.t. $\| X \|_{\infty,2} := \mathbb{E}[\sup_{0\leq t\leq T}|X_t|^2]<\infty$} and there exists a constant $C$, which depends only on $T$ and $L$ such that

$$\mathbb{E} \bigg[\sup_{0\leq t\leq T} |X_t|^2 \bigg] \leq C \mathbb{E} \bigg[|x_0|^2 + \int_0^T\bigg( |b(t,0)|^2 + |\sigma(t,0)|^2 \bigg) dt \bigg]$$

proof

The proof is divided into two parts, first he shows that $$\mathbb{E} \bigg[\sup_{0\leq t\leq T} |X_t|^2 \bigg] \leq C \mathbb{E} \bigg[|x_0|^2 + \int_0^T\bigg( |b(t,0)|^2 + |\sigma(t,0)|^2 \bigg) dt \bigg] + C \mathbb{E} \bigg[\int_0^T |X_t|^2 dt\bigg]<\infty \quad (*)$$ Then he shows that for any $\varepsilon>0$, there exists a constant $C_\varepsilon>0$ such that $$\sup_{0\leq t \leq T} \mathbb{E} \bigg[|X_t|^2 \bigg] \leq \varepsilon \mathbb{E}\bigg[\sup_{0\leq T} |X_t|^2 \bigg] + C_\varepsilon \mathbb{E} \bigg[|x_0|^2 + \int_0^T\bigg( |b(t,0)|^2 + |\sigma(t,0)|^2 \bigg) dt \bigg] \quad (**)$$

My question is, how does plugging the $(**)$ in $(*)$ yields the result?

Answer 1

The equality $$ \sup_{0\leq t \leq T} \mathbb{E} \big[|X_t|^2 \big] = \mathbb{E} \Big[\sup_{0\leq t\leq T} |X_t|^2 \Big] $$ is a direct consequence of the theorem of monontone convergence for the integral $\int_\Omega Y\,d\mathbb P=\mathbb E[Y]\,.$ The equality $$ \mathbb{E} \Big[\sup_{0\leq t \leq T} |X_t|^2 \Big] = \mathbb{E} \Big[\Big(\sup_{0\leq t\leq T} |X_t|\Big)^2 \Big] $$ follows from $$\tag{1} \sup_{0\leq t \leq T} |X_t|^2 = \Big(\sup_{0\leq t\leq T} |X_t|\Big)^2\quad\text{ a.s. } $$ Proof of (1). Because the process $X$ is a.s. continuous the left hand side of (1) is a.s. finite. So is $\sup_{0\leq t \leq T} |X_t|\,.$ Pick $\omega\in\Omega$ for which $t\mapsto X_t(\omega)$ is continuous and write $$ x_m=\sup_{0\leq t \leq T} |X_t|(\omega)\,. $$ On the compact interval $[0,x_m]$ the function $f(x)=x^2$ attains its maximum in $x_m\,.$ In particular, $$ f(x_m)=\sup_{x\in[0,x_m]}f(x)\,. $$ In other words, $$ \Big(\sup_{0\le t\le T}|X_t|(\omega)\Big)^2=\sup_{0\le t\le T}|X_t|^2(\omega)\,. $$